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Maksim231197 [3]
2 years ago
9

How would you obtain a mean value for the specific heat capacity of a material?

Physics
1 answer:
lidiya [134]2 years ago
8 0
The heat capacity and the specific heat are related by C=cm or c=C/m. The mass m, specific heat c, change in temperature ΔT, and heat added (or subtracted) Q are related by the equation: Q=mcΔT. Values of specific heat are dependent on the properties and phase of a given substance.
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What was done to test if Comets are responsible for 1/2 the Earth's water?
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✯Hello✯

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Difference between duracell and interstate golf cart batteries
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3 0
3 years ago
A uniform meterstick of mass 0.20 kg is pivoted at the 40 cm mark. where should one hang a mass of 0.50 kg to balance the stick?
Tcecarenko [31]
The weight of the meterstick is:
W=mg=0.20 kg \cdot 9.81 m/s^2 = 1.97 N
and this weight is applied at the center of mass of the meterstick, so at x=0.50 m, therefore at a distance 
d_1 = 0.50 m - 0.40 m=0.10 m
from the pivot.
The torque generated by the weight of the meterstick around the pivot is:
M_w = W d_1 = (1.97 N)(0.10 m)=0.20 Nm

To keep the system in equilibrium, the mass of 0.50 kg must generate an equal torque with opposite direction of rotation, so it must be located at a distance d2 somewhere between x=0 and x=0.40 m. The magnitude of the torque should be the same, 0.20 Nm, and so we have:
(mg) d_2 = 0.20 Nm
from which we find the value of d2:
d_2 =  \frac{0.20 Nm}{mg}= \frac{0.20 Nm}{(0.5 kg)(9.81 m/s^2)}=0.04 m

So, the mass should be put at x=-0.04 m from the pivot, therefore at the x=36 cm mark.
4 0
3 years ago
A spherical soap bubble with a surface-tension of 0.005 lbf/ft is expanded from a diameter of 0.5 in to 3.0 in. How much work, i
Sladkaya [172]

Answer:W = 1.23×10^-6BTU

Explanation: Work = Surface tension × (A1 - A2)

W= Surface tension × 3.142 ×(D1^2 - D2^2)

Where A1= Initial surface area

A2= final surface area

Given:

D1=0.5 inches , D2= 3 inches

D1= 0.5 × (1ft/12inches)

D1= 0.0417 ft

D2= 3 ×(1ft/12inches)

D2= 0.25ft

Surface tension = 0.005lb ft^-1

W = [(0.25)^2 - (0.0417)^2]

W = 954 ×10^6lbf ft × ( 1BTU/778lbf ft)

W = 1.23×10^-6BTU

8 0
3 years ago
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